JEE Main 2026 · 2 Apr · Shift 2 · Chemistry Q68

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ChemistrySingle correct+4 / -1Question 68 of 75

Identify compounds A and E in the following reaction sequence.

1-Ethyl-4-nitrobenzene → (Br₂/AlBr₃) A → (Sn/HCl) B → (NaNO₂/HCl, 273-278 K) C → (C₂H₅OH) D → ((i) KMnO₄/KOH; (ii) H₃O⁺) E.

A = 2-bromo-4-ethyl-1-nitrobenzene (Br ortho to NO₂); E = 3-bromobenzoic acid.

A = 2-bromo-1-ethyl-4-nitrobenzene (Br ortho to the ethyl group); E = 2-bromobenzoic acid.

A = 1-(2-bromoethyl)-4-nitrobenzene; E = benzoic acid.

A = 2-bromo-1-ethyl-4-nitrobenzene; E = 2-bromo-4-hydroxybenzoic acid.

Answer and worked solution

Official answer: B

Bromination. In 1-ethyl-4-nitrobenzene, ethyl (o/p director) and NO₂ (meta director) both direct Br to the position ortho to ethyl. A = 2-bromo-1-ethyl-4-nitrobenzene.

Remove the nitrogen.

Sn/HCl reduces $\ce{NO2}$ to $\ce{NH2}$ (B).
Diazotisation gives C.
Ethanol replaces $\ce{N2+}$ by H (D = 1-bromo-2-ethylbenzene).

Oxidation. Alkaline $\ce{KMnO4}$ oxidises the ethyl side chain to COOH: E = 2-bromobenzoic acid.

(B)

Topics:Diazonium salts Aromatic hydrocarbons and electrophilic substitution Oxidation and reduction

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JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 68